A063250 Number of binary right-rotations (iterations of A038572) to reach fixed point.
0, 0, 1, 0, 2, 2, 1, 0, 3, 3, 3, 3, 2, 2, 1, 0, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 0, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 0, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 0
Examples
a(11)=3 since under right-rotation 11 -> 13 -> 14 -> 7 and 7 is a fixed point.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..32767
Programs
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Mathematica
Table[Length[FixedPointList[FromDigits[RotateRight[IntegerDigits[ #,2]], 2]&, n]]-2,{n,0,110}] (* Harvey P. Dale, Dec 23 2011 *)
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Python
def a(n): if n<2: return 0 b=bin(n)[2:] s=0 while "0" in b: N=int(b[-1] + b[:-1], 2) s+=1 b=bin(N)[2:] return s print([a(n) for n in range(105)]) # Indranil Ghosh, May 25 2017
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Python
def A063250(n): return 0 if (t:=bin(n)[2:].find('0'))==-1 else n.bit_length()-t # Chai Wah Wu, Mar 09 2025
Formula
If n+1 is a power of 2 then a(n)=0 otherwise a(n) = 1 + a(floor(n/2)).
Conjectured g.f.: 1/(1-x) * Sum_{j>=0} x^(2^j) - (1-x^(2^j)) * Sum_{k>=1} x^((2^j)*(2^k-1)). - Mikhail Kurkov, Sep 29 2019
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